Chap17.4- Chap17.5
Applied Mechanics II
EQUATIONS OF MOTION:
ROTATION ABOUT A FIXED AXIS
Today’s Objectives:
Students will be able to:
1. Analyze the planar kinetics
of a rigid body undergoing
rotational motion.
In-Class Activities:
• Rotation about an Axis
• Equations of Motion
2
APPLICATIONS
The crank on the oil-pump rig undergoes
rotation about a fixed axis, caused by the
driving torque, M, from a motor.
As the crank turns, a dynamic reaction is
produced at the pin. This reaction is a
function of angular velocity, angular
acceleration, and the orientation of the
crank.
Pin at the center of
rotation.
If the motor exerts a constant torque M
on the crank, does the crank turn at a
constant angular velocity? Is this
3
desirable for such a machine?
APPLICATIONS (continued)
The pendulum of the Charpy
impact machine is released from
rest when  = 0°. Its angular
velocity () begins to increase.
Can we determine the angular
velocity when it is in vertical
position?
On which property (P) of the
pendulum does the angular
acceleration () depend?
What is the relationship between4 P
EQUATIONS OF MOTION:
ROTATION ABOUT A FIXED AXIS
(Section 17.4)
When a rigid body rotates about a fixed axis
perpendicular to the plane of the body at
point O, the body’s center of gravity G moves
in a circular path of radius rG. Thus, the
acceleration of point G can be represented by
a tangential component (aG)t = rG  and a
normal component (aG)n = rG 2.
Since the body experiences an angular acceleration, its inertia
creates a moment of magnitude, Ig, equal to the moment of the
external forces about point G. Thus, the scalar equations of
motion can be stated as:
 Fn = m (aG)n = m rG 2
 Ft = m (aG)t = m rG 
5
 MG = IG 
EQUATIONS OF MOTION
(continued)
Note that the MG moment equation may be replaced by a
moment summation about any arbitrary point. Summing the
moment about the center of rotation O yields
MO = IG + rG m (aG)t = [IG + m(rG)2] 
From the parallel axis theorem,
IO = IG + m(rG)2, therefore the term
in parentheses represents IO.
Consequently, we can write the three equations of
motion for the body as:
Fn = m (aG)n = m rG 2
Ft = m (aG)t = m rG 
MO = IO 
6
PROCEDURE FOR ANALYSIS
Problems involving the kinetics of a rigid body rotating about
a fixed axis can be solved using the following process.
1. Establish an inertial coordinate system and specify the sign and
direction of (aG)n and (aG)t.
2. Draw a free body diagram accounting for all external forces and couples. Show the
resulting inertia forces and couple (typically on a separate kinetic diagram).
3. Compute the mass moment of inertia IG or IO.
4. Write the three equations of motion and identify the unknowns. Solve for the
unknowns.
5. Use kinematics if there are more than three unknowns (since the equations of
motion allow for only three unknowns).
7
◼Free body diagram = kinetic diagram
FBD
Kinetic Diagram
IG 
rG= 0.15 m
mg
On
•
=
Ot
 Fx = m (aG)x
 Fy = m (aG)y
 MG = I G 
•
m rG 2 = 0
8
m rG 
or
 Fx = m (aG)x
 Fy = m (aG)y
 MO = IO 
Depending on
which location
to take the
moment
balance, the
mass moment
inertia is
different
EXAMPLE I
Given:A rod with mass of 20
kg is rotating at 5 rad/s
at the instant shown.
A moment of 60 N·m
is applied to the rod.
Find: The angular acceleration  and the reaction at pin O
when the rod is in the horizontal position.
Plan: Since the mass center moves in a circle of radius
1.5 m, it’s acceleration has a normal component
toward O and a tangential component acting
downward and perpendicular to rG.
Apply the problem solving procedure.
9
EXAMPLE I (continued)
Solution:
FBD
Kinetic Diagram
=
Equations of motion:
+ → Fn = m an = m rG 2
On = 20(1.5)(5)2 = 750 N
+  Ft = m at = m rG 
-Ot + 20(9.81) = 20(1.5) 
10
EXAMPLE I (continued)
Solution:
FBD
Kinetic Diagram
=
+ MO = IG  + m (rG ) (rG) =
 1.5 (20) 9.81+60 = IG  + m(rG)2 = IO 
L
Using IG = (ml2)/12 and rG = (0.5)(l), we can write:
MO = [(ml2/12) + (ml2/4)]  = (ml2/3)  where (ml2/3) = IO.
IO
After substituting:
60 + 20(9.81)(1.5) = 20(32/3) 
Solving:  = 5.9 rad/s2
Ot = 19 N
11
Example 17.8
Release from rest, so
at that instant, no
angular velocity.
Because of moment
imbalance, flywheel is
subject to angular
acceleration
12
Example 17.8
Release from rest, so
at that instant, no
angular velocity.
Because of moment
imbalance, flywheel is
subject to angular
acceleration
13
Example 17.8
Free body diagram
Kinetic diagram
14
Example 17.8
◼What if we Take moment with respect to O
α × rG =at
The net moment is equivalent to
Ig×α + m ×at
15
Example 17.8
◼What if we use parallel-axis theorem
16
Method1
17
Method2
18
Method3
Io α
19
Example 17.10
Example 17.10
◼First thing is to draw free body diagram and
kinetic diagram
Not all the unknowns are
21
independent
Example 17.10
T (tension of the rope)
Is involved
22
Example 17.10
23
Example 17.10 (method2)
◼Without get T (tension of the rope) involved
24
Example 17.11
25
Example 17.11
an
α
at
26
Example 17.11
◼At any angle θ, the equations of motion is the
following:
Mg Sinθ
Mg Cosθ
Free body diagram
kinetic diagram
Note that
mass
moment
inertia is w.r.t.
27 A
location
Example 17.11
◼Although there are 4 unknown, α and ω are
not independent
Mg Sinθ
Mg Cosθ
Free body diagram
kinetic diagram
28
Example 17.11
29
EXAMPLE II
Given:The uniform slender rod has a
mass of 15 kg and its mass
center is at point G.
•G
Find: The reactions at the pin O and
the angular acceleration of the
rod just after the cord is cut.
Plan: Since the mass center, G, moves in a circle of radius
0.15 m, it’s acceleration has a normal component toward
O and a tangential component acting downward and
perpendicular to rG.
Apply the problem solving procedure.
30
EXAMPLE II (continued)
Solution:
FBD
Kinetic Diagram
IG 
rG= 0.15 m
mg
On
•
=
Ot
•
m rG 2 = 0
m rG 
Equations of motion:
+ Fn = m an = m rG 2
 On = 0 N
+ Ft = m at = m rG 
 -Ot + 15(9.81) = 15 (0.15)  (1)
=0
m rG 
+ MO = IG  + m rG  (rG)  0.15 (15) 9.81 = IG  + m(rG)2 
Using IG = (ml2)/12 and rG = (0.15), we can write:
IG  + m(rG)2  = [(15×0.92)/12 + 15(0.15)2]  = 1.35 
31
EXAMPLE II (continued)
FBD
Kinetic Diagram
mg
IG 
rG= 0.15 m
On
•
•
=
Ot
m rG 2 = 0
m rG 
After substituting:
22.07 = 1.35    = 16.4 rad/s2
From Eq (1) :
-Ot + 15(9.81) = 15(0.15)
 Ot = 15(9.81) − 15(0.15)16.4 = 110 N
32
GROUP PROBLEM SOLVING I
Given: The 4-kg slender rod is
initially supported
horizontally by a spring
at B and pin at A.
Find: The angular acceleration of the rod and the acceleration
of the rod’s mass center at the instant the 100-N force is
applied.
Plan:
Find the spring reaction force before the 100 N is
applied. Draw the free body diagram and kinetic
diagram of the rod. Then apply the equations of
motion.
33
GROUP PROBLEM SOLVING I
Solution:
FBD
Kinetic Diagram
IG 
An
At
4(9.81) N
Rsp= 19.62 N
m(1.5)2 = 0
m(1.5)
Notice that the spring force, Rsp developed before the application of the 100 N force is
half of the rod weight:
Rsp = 4 (9.81) / 2 = 19.62 N
Equation of motion:
+ MA = IG  + m rG  (rG)
 - 19.62(3) + 100(1.5) + 4(9.81)(1.5) = IG  + m(rG)2 
34
GROUP PROBLEM SOLVING I
(continued)
FBD
Kinetic Diagram
IG 
An
At
4(9.81) N
Rsp= 19.62 N
m(1.5)2 = 0
m(1.5)
Using IG = (ml2)/12 and rG = (1.5), we can write:
IG  + m(rG)2  = [(4×32)/12 + 4(1.5)2]  = 12 
After substituting:
150 = 12    = 12.5 rad/s2
The acceleration of the rod’s mass center is :
an = rG 2 = 0 m/s2
at = rG  = 18.8 m/s2 
35
READING QUIZ
1. In rotational motion, the normal component of acceleration
at the body’s center of gravity (G) is always __________.
A) zero
B) tangent to the path of motion of G
C) directed from G toward the center of rotation
D) directed from the center of rotation toward G
2. If a rigid body rotates about point O, the sum of the
moments of the external forces acting on the body about
point O equals which of the following?
A) IG 
B) IO 
C) m aG
D) m aO
m
m
ATTENTION QUIZ
1. A drum of mass m is set into motion in two
ways: (a) by a constant 40 N force, and, (b) by a
block of weight 40 N. If a and b represent the
angular acceleration of the drum in each case,
select the true statement.
A) a > b
C) a = b
T
(a)
B) a < b
D) None of the above
2. In case (b), what is the tension T in the cable?
A) T = 40 N
B) T < 40 N
C) T > 40 N
D) None of the above
(b)
CONCEPT QUIZ
1. If a rigid bar of length l (above) is released
from rest in the horizontal position ( = 0),
the magnitude of its angular acceleration is
at maximum when
A)  = 0
B)  = 90
C)  = 180
D)  = 0 and 180
2. In the above problem, when  = 90°, the horizontal
component of the reaction at pin O is __________.
A) zero
B) m g
C) m (l/2) 2
D) None of the above
17.5 EQUATIONS OF MOTION:
GENERAL PLANE MOTION
Today’s Objectives:
Students will be able to:
1. Analyze the planar kinetics of a
rigid body undergoing general
plane motion.
In-Class Activities:
• Equations of Motion
• Frictional Rolling Problems
39
APPLICATIONS
As the soil compactor accelerates
forward, the front roller experiences
general plane motion (both translation
and rotation).
How would you find the loads
experienced by the roller shaft or its
bearings?
=
The forces shown on the
roller shaft’s FBD cause the
accelerations shown on the
kinetic diagram.
Is point A the IC?
40
APPLICATIONS (continued)
The lawn roller is pushed forward with a force of 200 N when
the handle is held at 45°.
How can we determine its translational acceleration and
angular acceleration?
Does the total acceleration depend on the coefficient’s of static
and kinetic friction?
41
APPLICATIONS (continued)
During an impact, the center of
gravity G of this crash dummy will
decelerate with the vehicle, but also
experience another acceleration due
to its rotation about point A.
Why?
How can engineers use this information to determine the forces
exerted by the seat belt on a passenger during a crash? How
would these accelerations impact the design of the seat belt
itself?
42
EQUATIONS OF MOTION:
GENERAL PLANE MOTION (Section
17.5)
When a rigid body is subjected to external
forces and couple-moments, it can undergo
both translational motion and rotational
motion. This combination is called general
plane motion.
Using an x-y inertial coordinate system,
the scalar equations of motions about the
center of mass, G, may be written as:
 Fx = m (aG)x
 Fy = m (aG)y
 MG = IG 
43
EQUATIONS OF MOTION:
GENERAL PLANE MOTION
(continued)
Sometimes, it may be convenient to write the
moment equation about a point P, rather than
G. Then the equations of motion are written
as follows:
 Fx = m (aG)x
 Fy = m (aG)y
 MP =  (Mk )P
In this case,  (Mk )P represents the sum of the moments of
IG and maG about point P.
44
FRICTIONAL ROLLING
PROBLEMS
When analyzing the rolling motion of wheels, cylinders, or disks,
it may not be known if the body rolls without slipping or if it
slips/slides as it rolls.
For example, consider a disk with mass m
and radius r, subjected to a known force P.
The equations of motion will be:
 Fx = m (aG)x  P − F = m aG
 Fy = m (aG)y  N − mg = 0
 MG = I G   F r = I G 
There are 4 unknowns (F, N, , and aG) in
45
these three equations.
FRICTIONAL ROLLING
PROBLEMS (continued)
Hence, we have to make an assumption
to provide another equation. Then, we
can solve for the unknowns.
The 4th equation can be obtained from
the slip or non-slip condition of the disk.
Case 1:
Assume no slipping and use aG =  r as the 4th equation and DO NOT use Ff = sN.
After solving, you will need to verify that the assumption was correct by checking if
Ff  sN.
Case 2:
Assume slipping and use Ff = kN as the 4th equation.
In this case, aG  r.
46
◼Non-slipping: the point C and the point G
have the same
c
47
PROCEDURE FOR ANALYSIS
Problems involving the kinetics of a rigid body undergoing
general plane motion can be solved using the following procedure.
1. Establish the x-y inertial coordinate system. Draw both the
free body diagram and kinetic diagram for the body.
2. Specify the direction and sense of the acceleration of the mass center, aG, and
the angular acceleration  of the body. If necessary, compute the body’s mass
moment of inertia IG.
3. If the moment equation Mp= (Mk)p is used, use the kinetic diagram to help
visualize the moments developed by the components m(aG)x, m(aG)y, and IG.
4. Apply the three equations of motion.
48
PROCEDURE FOR ANALYSIS
(continued)
5. Identify the unknowns. If necessary (i.e., there are four
unknowns), make your slip-no slip assumption (typically no
slipping, or the use of aG =  r, is assumed first).
6. Use kinematic equations as necessary to complete the solution.
7. If a slip-no slip assumption was made, check its validity!!!
Key points to consider:
1. Be consistent in using the assumed directions.
The direction of aG must be consistent with .
2. If Ff = kN is used, Ff must oppose the motion. As a test,
assume no friction and observe the resulting motion.
This may help visualize the correct direction of Ff.
49
Example 17.12
50
Example 17.12
51
Example 17.12
52
Example 17.12 (method 2)
53
Example 17.12 (method 3)
54
Parallel-axis theorem
Example 17.14
55
Example 17.14
an is zero
because
there is no
initial
angular
velocity
Free body diagram
kinetic diagram
56
Example 17.14
an is zero
because
there is no
initial
angular
velocity
Free body diagram
kinetic diagram
57
Example 17.14 (without slipping)
58
Example 17.14 (with slipping)
59
Example 17.5
60
Example 17.5
Note (aG)x is zero
because when the cord
AC is cut, the instant
velocity at G is zero.
61
Example 17.5
62
Example 17.5
63
EXAMPLE
Given: A spool has a mass of 200 kg and a radius of gyration (kG)
of 0.3 m. The coefficient of kinetic friction between the
spool and the ground is k = 0.1.
Find: The angular acceleration () of the spool and the tension
in the cable.
Plan: Focus on the spool. Follow the solution procedure (draw
64
a FBD, etc.) and identify the unknowns.
EXAMPLE (continued)
Solution:
The free-body diagram and kinetic diagram for the body are:
IG 
=
maG
1962 N
Equation of motion in the y-direction (do first since there is only one
unknown):
+ Fy = m (aG)y : NB − 1962 = 0
 NB = 1962 N
No angular velocity,
thus, (aG)y is zero
65
EXAMPLE (continued)
IG 
=
maG
1962 N
Note that aG = (0.4) . Why???????? The spool rotates with respect to A. A is IC
(instantaneous center)
+→ Fx = m (aG)x: T – 0.1 NB = 200 aG = 200 (0.4) 
 T – 196.2 = 80 
+ MG = IG  : 450 – T(0.4) – 0.1 NB (0.6) = 20 (0.3)2 
 450 – T(0.4) – 196.2 (0.6) = 1.8 
Solving these two equations, we get
 = 7.50 rad/s2, T = 797 N
66
GROUP PROBLEM SOLVING
Given: The 500-kg
concrete culvert has
a mean radius of
0.5 m. Assume the
culvert does not
slip on the truck bed
but can roll, and you
can neglect its
thickness.
Find: The culvert’s angular acceleration when the truck has
an acceleration of 3 m/s2.
Plan: Follow the problem-solving procedure.
67
GROUP PROBLEM SOLVING
(continued)
Solution:
The moment of inertia of the culvert about G is
IG = m(r)2 = (500)(0.5)2 = 125 kg·m2
Draw the FBD
and
Kinetic Diagram.
y
500(9.81) N
x
=
G
125
G
500 aG
0.5m
0.5m
A
F
N
A
68
GROUP PROBLEM SOLVING
(continued)
y
500(9.81) N
x
=
G
0.5m
A
F
125
G
500 aG
0.5m
A
N
Equations of motion: the moment equation of motion about A
+ MA = (Mk)A
0 = 125  – 500 aG (0.5)
(1)
69
GROUP PROBLEM SOLVING
(continued)
Since the culvert does not slip at A,
aA = 3 m/s2.
Apply the relative acceleration equation to find
 and aG.
y
, 
x
aG
G
aG = aA +   rG/A − 2 rG/A
2(0.5)
aG i = 3 i +  k  0.5 j −
= (3 −0.5 ) i − 2(0.5) j
0.5m
j
(aA)t
A
(aA)n
Equating the i components,
aG = (3 − 0.5 )
(2)
Solving Equations (1) and (2) yields
 = 3 rad/s2 and aG = 1.5 m/s2 →
70
CONCEPT QUIZ

1. An 80 kg spool (kG = 0.3 m) is on a
rough surface and a cable exerts a 30 N
0.2m
load to the right. The friction force at A
•G
acts to the __________ and the aG
0.75m
should be directed to the __________ .
A
A) right, left
B) left, right
C) right, right D) left, left
30N
2. For the situation above, the moment equation about G is?
A) 0.75 (FfA) - 0.2(30) = - (80)(0.32)
B) -0.2(30) = - (80)(0.32)
C) 0.75 (FfA) - 0.2(30) = - (80)(0.32) + 80aG
D) None of the above
READING QUIZ
1. If a disk rolls on a rough surface without slipping, the
acceleration of the center of gravity (G) will _________ and
the friction force will be __________.
A) not be equal to  r; less than sN
B) be equal to  r; equal to kN
C) be equal to  r; less than sN
D) None of the above
2. If a rigid body experiences general plane motion, the sum of
the moments of external forces acting on the body about any
point P is equal to __________.
A) IP 
B) IP  + maP
C) m aG
D) IG  + rGP × maP
ATTENTION QUIZ
1. A slender 100 kg beam is suspended by a
cable. The moment equation about point A is?
A)
B)
C)
D)
3(10) = 1/12(100)(42) 
3(10) = 1/3(100)(42) 
3(10) = 1/12(100)(42)  + (100 aGx)(2)
None of the above
A
3m
10 N
2. Select the equation that best represents the “no-slip”
assumption.
A) Ff = s N
B) Ff = k N
C) aG = r 
D) None of the above
4m